Move Last Element to Front of a Given Linked List
Problem statement: Given a Singly linked list, move the last element to the front of the linked list.
For example,
-> 1 -> 2 -> 3 -> 4 -> 5 -> NULL -> 5 -> 1 -> 2 -> 3 -> 4 -> NULL |
Here we traverse the linked list till the next pointer is NULL, store the last and second last element in two variables.
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NODE *secLast = NULL;
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NODE *last = head;
  while
(last->next != NULL)
   Â
{
       Â
secLast = last;
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last = last->next;
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}
When we reach NULL pointer, assign the second last element’s next to NULL pointer.
   Â
secLast->next = NULL;
Assign last element’s next to head.
   Â
last->next = head;
At last assign the head to last element.
  Â
 head = last;
/*C code to move the last element to front of the Linked List */ #include <stdio.h> #include <stdlib.h> typedef struct node { int data; struct node *next; } NODE; NODE *head = NULL; NODE *newNode(int key) { NODE *temp = (NODE *)malloc(sizeof(NODE)); temp->data = key; temp->next = NULL; return temp; } void moveToFront() { if (head == NULL || (head)->next == NULL) return; NODE *secLast = NULL; NODE *last = head; while (last->next != NULL) { secLast = last; last = last->next; } secLast->next = NULL; last->next = head; head = last; } void printList() { NODE *temp = head; if (temp == NULL) { printf("List is empty!\n"); return; } printf("Linked List is\t"); while (temp != NULL) { printf("%d--> ", temp->data); temp = temp->next; } printf("NULL\n"); } int main() { head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(4); head->next->next->next->next = newNode(5); head->next->next->next->next->next = NULL; printList(); moveToFront(); printf("Linked list after moving last element to front\n"); printList(); return 0; }
$ ./a.out Linked List is 1–> 2–> 3–> 4–> 5–> NULL Linked list after moving last element to front Linked List is 5–> 1–> 2–> 3–> 4–> NULL $ |
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